2788 + Interview Questions in GRAVITATION IN GENERAL AWARENESS Page 55 InterviewSolution

2701.	A person brings a mass of 1kg from infinity to a point A. Initially the mass was at rest but it moves at a speed of 2m/s as it reaches A. The work done by the person on the mass is —3J. The potential at A is(a) -3J/kg (b) -2J/kg (c) -5JAg (d) none of these.
Answer» The correct answer is (c) -5 J/kg EXPLANATION: Work done by the person = -3 J (Given) Work done in increasing the velocity from rest to 2 km/s = K.E. of the mass at A =½mv² = ½12² = 2 J Let the potential at A be X J/kg, So, the change in P.E. of the mass = the work done in slowly bringing the mass from infinity to A = mX = 1*X = X J So the total work done = Change in P.E + Change in K.E. = X+2 J, but it is given -3 J so, X+2 = -3 →X = -3-2 = -5 J

2701.

A person brings a mass of 1kg from infinity to a point A. Initially the mass was at rest but it moves at a speed of 2m/s as it reaches A. The work done by the person on the mass is —3J. The potential at A is(a) -3J/kg (b) -2J/kg (c) -5JAg (d) none of these.

Answer»

The correct answer is (c) -5 J/kg

EXPLANATION:

Work done by the person = -3 J (Given) Work done in increasing the velocity from rest to 2 km/s = K.E. of the mass at A =½mv² = ½*1*2² = 2 J Let the potential at A be X J/kg,

So, the change in P.E. of the mass = the work done in slowly bringing the mass from infinity to A = mX

= 1*X = X J

So the total work done = Change in P.E + Change in K.E. = X+2 J, but it is given -3 J so, X+2 = -3

→X = -3-2 = -5 J

Discussion

2702.	A satellite can be in a geostationary orbit around earth at a distance r from the centrer. If the angular velocity of earth about its axis doubles, a satellite can now be in a geostationary orbit around if its distance from the centre isA. `(r )/(2)`B. `(r )/(2sqrt(2))`C. `(r )/((4)^(1//3))`D. `(r )/((2)^(1//3))`
Answer» Correct Answer - C Let angular velocity is `omega rArr mr omega^(2)=(GMm)/(r^(2))` `omega^(2)=(GM)/(r^(3))` so `omega_(1)^(2)r_(1)^(3)=omega_(2)^(2)r_(2)^(3)` `r_(2)^(3)=r_(1)^(3)((omega_(1))/(omega_(2)))^(2) rArr r_(2)^(3)=(r_(1)^(3))/(4) rArr r_(2)=(r_(1))/(4^(1//3))`

Discussion

2703.	A person brings a mass of 1 kg from infinity to a point . Initally the mas was at rest but it moves at a speed of 2 `ms^-1` as it reaches A. The work done by the perosn on the mass is -3J. The potential at A isA. `-3Jkg^-1`B. `-2Jkg^-1`C. `-5Jkg^-1`D. none of these
Answer» Correct Answer - C

Discussion

2704.	A person brings a mass of 1 kg from infinity to a point . Initally the mass was at rest but it moves at a speed of 2 `ms^-1` as it reaches A. The work done by the person on the mass is -3J. The potential at A isA. `-3 "J kg"^(-1)`B. `-12 "J kg"^(-1)`C. `-5 "J kg"^(-1)`D. None of these
Answer» Correct Answer - C `W=E_(A)-E_(infty)=(U_(A)+K_(A))-(U_(infty)+K_(infty))` `:. -3={(1)V_(A)+(1)/(2)xx(1)(2)^(2)}-(0+0)" "(because W-3J)` `:. V_(A)=-5"J kg"^(-1)`.

Discussion

2705.	A satallite of mass `m`, initally at rest on the earth, is launched into a circular orbit at a height equal to the the radius of the earth. The minimum energy required isA. `(sqrt(3))/4 mgR`B. `1/2 mgR`C. `1/4 mgR`D. `3/4 mgR`
Answer» Correct Answer - d We know `V_(0)=sqrt((GM)/r) & g=(GM)/(R^(2))` From energy conservation `U_(i)+K_(i)=U_(f)+K_(f)` `-(GMm)/(2R)+1/2m(sqrt((GM)/(2R)))^(2)` `K_(i)=(3GMm)/(4R)impliesK_(i)=3/4 mgR`

Discussion

2706.	A satallite of mass `m`, initally at rest on the earth, is launched into a circular orbit at a height equal to the the radius of the earth. The minimum energy required isA. `(sqrt(3))/(4)mgR`B. `(1)/(2)mgR`C. `(1)/(4)mgR`D. `(3)/(4)mgR`
Answer» Correct Answer - D We know `V_(0) = sqrt((GM)/(r )) & g = (GM)/(R^(2))` From energy conservation `U_(i)+K_(i)=U_(f)+K_(f)` `-(GMm)/(R )+K_(f)=-(GMm)/(2R)+(1)/(2)mv_(0)^(1)` `K_(i)=(GMm)/(2R)+(1)/(2)m(sqrt((GM)/(2R)))^(2) rArr K_(i) = (3GMm)/(4R) rarr K_(i)=(3)/(4) mgR`

Discussion

2707.	A communications Earth satalliteA. goes round the earth from east to westB. can be in the equatorial plane onlyC. can be vertically any place on the earthD. goes round the earth from west to east
Answer» Correct Answer - B::D

Discussion

2708.	A bullet is fired vertically upwards with velocity v form the surface of a spherical planet. If the escape velocity from the planet is `v_(esc) = vsqrtN`, then the value of N is (ignore energy loss due to atmosphere)
Answer» Correct Answer - B Let h be the height to which the bullet rises `then, g^1 =g((1+(h)/(R ))^(-2)` `rArr (g)/(4) = g((1 +(h)/(R )^(-2)` `rArr h =R` we know that `v_e = sqrt((2GM)/(R ) = vsqrtN (given) …(i)` Now applying conservation of energy for the throw loss of kinetic energy = Gain in gravitational potential energy `:. (1)/(2)mv^2 =- (GMm)/(2R) - (-(GMm)/(R )` `:. v = sqrt((GM)/(R ) ...(ii)` Comparing (i) & (ii) N = 2

Discussion

2709.	Assertion: The total energy of a satellite is negative. Reason: Gravitational potential energy of an object is negative.A. If both the assertion and reason are true and reason is a true explantion of the assertion.B. If both the assertion and reason are true but the reason is not true the correct explantion of the assertion.C. If the assertion is true but reason falseD. If both the assertion and reason are false.
Answer» Correct Answer - A Total mechanical energy of the satellite is the sum of kinetic energy which is always positive and the gravitational potential energy which is negative and in magnitude the `K.E.` is half the `P.E.` So the the total energy of the satellite is negative. The total energy is negative for any bounded system, that is one in which the object is closest.

Discussion

2710.	A communications Earth satalliteA. goes round the earth from east to westB. can be in the equatorial plane onlyC. can be vertically above any place on the earthD. goes round the earth from west to east
Answer» Correct Answer - b,d

Discussion

2711.	Two particles of masses `1kg` and `2kg` are placed at a distance of `50cm`. Find the initial acceleration of the first particle due to gravitational force.
Answer» Gravitational force between two particles `F=(Gm_(1)m_(2))/(r^(2))=(6.67xx10^(-11)xx1xx2)/((0.5)^(2))=5.3xx10^(-10)N` The acceleration of `1Kg` particles is `a_(1)=F/(m_(1))=(5.3xx10^(-10))/1=5.3xx10^(-10)ms^(-2)` towards the `2Kg` mass

Discussion

2712.	A shell is fired vertically from the earth with speed `v_(esc)//N`, where `N` is some number greater than one and `v_(esc)` is escape speed on the earth. Neglecting the rotation of the earth and air resistance, the maximum altitude attained by the shell will be (`R_(E)` is radius of the earth):A. `(R_(E))/(N^(2)-1)`B. `(R_(E))/(N^(2))`C. `(NR_(E))/(N^(2)-1)`D. `(N^(2)R_(E))/(N^(2)-1)`
Answer» Correct Answer - A `-(GMm)/R=-(GMm)/(R+h)+1/2mv^(2)` `1/2v^(2)=GM(1/(R+h)-1/R)v^(2)=(2GM)/Rh/(R+h)` `v^(2)=v_(e)^(2)h/(R+h)` given `v=(v_(e))/N` Substitute `rArr h=R/(1-N^(2))`

Discussion

2713.	A particle of mass `M` is at a distance a from surface of a thin spherical shell of equal mass and having radius `a`. A. Gravitational field and potential both are zero at centre of the shell.B. Gravitational field is zero not only inside the shell but at a point outside the shell also.C. Inside the shell, gravitational field alone is zeroD. Neither gravitational field nor gravitational potential is zero inside the shell.
Answer» Correct Answer - d

Discussion

2714.	For particles of equal masses M that move along a circle of radius R under the action of their mutual gravitational attraction. Find the speed of each particle.
Answer» Correct Answer - `sqrt((GM)/(R)((2sqrt(2)+1)/(4)))`

Discussion

2715.	Two particles of equal mass (m) each move in a circle of radius (r) under the action of their mutual gravitational attraction find the speed of each particle.
Answer» Correct Answer - `[v=sqrt((Gm)/(4R))]`

Discussion

2716.	The gravitational potential due to the Earth is minimum at (A) the centre of the Earth. (B) the surface of the Earth. (C) a points inside the Earth but not at its centre. (D) infinite distance.
Answer» Correct answer is (A) the centre of the Earth.

Discussion

2717.	State Kepler’s law equal of area.
Answer» The line that joins a planet and the Sun sweeps equal areas in equal intervals of time.

Discussion

2718.	The binding energy of a satellite revolving around planet in a circular orbit is 3 × 109 J. Its kinetic energy is ______ (A) 6 × 109 J (B) -3 × 109 J (C) -6 × 10+9 J (D) 3 × 10+9 J
Answer» Correct answer is (D) 3 × 10⁺⁹ J

Discussion

2719.	Two stallites A and B revolve round the same planet in coplanar circular orbits lying in the same plane. Their periods of revolutions are 1h and 8h, respectively. The radius of the orbit of A is `10^(4)`km. The speed of B relative to A when they are closed in `kmh^(-1)` isA. `10^(4)pi`B. `2xx10^(4)pi`C. `10^(4)pi//2`D. `4xx10^(4)pi`
Answer» Correct Answer - a `(r_(2))/(r_(1))=((T_(1))/(T_(1)))^(2//3)=((8)/(1))^(2//3)=4` `r_(2)=4xxr_(1)` `=4xx10^(4)km` `v_(1)=r_(1)omega_(1)=10^(4)xx(2pi)/(T_(1))` `v_(1)=(2pixx10^(4))/(1)=2pixx10^(4)` ltBrgt `v_(2)=v_(2)omega_(2)` `=4xx10^(4)xx(2pi)/(T_(2))` `=(4xx10^(4)xx2pi)/(8)=pixx10^(4)"km/h"` `V_(AB)=V_(B)-V_(A)` `=2pixx10^(4)-pixx10^(4)=pixx10^(4)(2-1)` `pixx10^(4)`.

Discussion

2720.	In case of an orbiting satellite, if the radius of orbit is decreasedA. its Kinetic Energy decreasesB. its Potential Energy decreasesC. its Mechanical Energy decreasesD. its speed decreases
Answer» Correct Answer - B::C

Discussion

2721.	How much energy is required by a satellite to keep it orbiting? Neglect air resistance, why?
Answer» No energy is required by a satellite to keep it orbiting. This is because the work done by the centripetal force is zero.

Discussion

2722.	Gravitational force between a point mass m and M separated by a distance is F. Now if a point mass 2m is placed next to m is contact with it. The force on M due to `m` and the total force on M areA. 2F, FB. F, 2FC. F, 3FD. F, F
Answer» Correct Answer - C `F=(Gm_(1)m_(2))/(r^(2))or F prop m_(1)m_(2)` On M due to m force is F and due to mass 2m force is 2F. Therefore, net force is 3F.

Discussion

2723.	The distance of planet Jupiter from the Sun is `5.2` times that of the earth. Find the period of revolution of Jupiter around the Sun.
Answer» Correct Answer - `11.86 years` Here, `R_(1)=R=`distance of earth from sun `R_(2)=5.2R T_(1)=1` year, `T_(2)=?` As `(T_(2)/T_(1))^(2)=(R_(2)/R^(1))^(3)=((5.2R)/(R))^(3)` `T_(2)=(5.2)^(3//2)xx1 year =5.2sqrt5.2 years` `T_(2)=11.86 years`

Discussion

2724.	Two particles of masses m and 3m are moving under their mutual gravitational force, around their centre of mass, in circular orbits. The separation between the masses is r. The gravitational attraction the two provides nessary centripetal force for circular motion The ratio of the potential energy to total kinetic energy of the system isA. `-2`B. `-1`C. 1D. 2
Answer» Correct Answer - A

Discussion

2725.	Two particles of masses m and 3m are moving under their mutual gravitational force, around their centre of mass, in circular orbits. The separation between the masses is r. The gravitational attraction the two provides nessary centripetal force for circular motion The ratio of centripetal forces acting on the two masses will beA. 1B. 2C. 3D. None of these
Answer» Correct Answer - A

Discussion

2726.	Gravitational force between two point masses `m` and `M` separated by a distance `r` is `F`. Now if a point mass `3m` is placed next to `m`, what will be that a. force on `M` due to `m`, b total on `M`?
Answer» a. If `r` is the distance between two point masses `m` and `M`, then the gravitational force on m due to mass `M` is `F=(GMm)/(r^(2))`. b. Total force on the body of mass `M` is `F`, i.e. `F=(GMxx(m+3m))/(r^(2))=(4GMm)/(r^(2))=4F`

Discussion

2727.	A ball is thrown vertically upwards with a velocity of `98m//s`. Calculate (i) The maximum height to which it rises. (ii) Total time it takes to return to earth.
Answer» Here, `u=98 m//s, v=0, a=-9.8M//S^(2)` `s=h=? T=?` From `v^(2)-u^(2)=2` as `0-(98)^(2)=2(-9.8)h` `h=(98xx98)/(2xx9.8)=490m` from `v=u-g t` `0=98-9.8t, t=98/9.8=10s` total time to return to earth, `2t=20 s`.

Discussion

2728.	How much will a body of mass once kg weigh on moon ? Given mass of moon is `(1)/(100)` mass of earth and diameter of moon is `(1)/(4)` the diameter of earth.
Answer» Here, `W_(m)=?m=(1)kg`. `M_(m)=(1)/(100)M_(e), D_(m)=(1)/(4)D_(e), R_(m)=(1)/(4)R_(e)` `W_(m)/W_(e)=M_(m)/R_(m)^(2).R_(e)^(2)/M_(e)=M_(m)/M_(e)xx(R_(e)/R_(m))^(2)=(1)/(100)xx(4)^(2)` `W_(m)=(16)/(100)W_(e)=(16)/(100)(mg_(e))=(16)/(100)xx1xx9.8N=1.57N`

Discussion

2729.	In the following figure, an orbit of a planet around the Sun (S) has been shown. AB and CD are the distances covered by the planet in equal time. Lines AS ad CS sweep equal areas in equal intervals of time. Hence, areas ASB and CSD are equal.(a) Which laws do we understand from the above description?(b) Write the law regarding the area swept.(c) Write the law T2 ∝ r3 in your words.(Schematic diagram)
Answer» (a) From the given description we understand Kepler’s three laws. (b) Kepler’s law of areas: The line joining the planet and the Sun sweeps equal areas in equal intervals of time. (c) Kepler’s law of periods: The square of the period of revolution of a planet around the Sun is directly proportional to the cube of the mean distance of the planet from the Sun.

2729.

In the following figure, an orbit of a planet around the Sun (S) has been shown. AB and CD are the distances covered by the planet in equal time. Lines AS ad CS sweep equal areas in equal intervals of time. Hence, areas ASB and CSD are equal.(a) Which laws do we understand from the above description?(b) Write the law regarding the area swept.(c) Write the law T2 ∝ r3 in your words.(Schematic diagram)

Answer»

(a) From the given description we understand Kepler’s three laws.

(b) Kepler’s law of areas: The line joining the planet and the Sun sweeps equal areas in equal intervals of time.

(c) Kepler’s law of periods: The square of the period of revolution of a planet around the Sun is directly proportional to the cube of the mean distance of the planet from the Sun.

Discussion

2730.	How much would a 70 kg man weigh on moon ?what will be his mass on earth and moon ? Given g on moon`=1.7 m//s^(2)`.
Answer» Correct Answer - `119 N, 70 kg, 70kg` `W_(m)=mg_(m)=70xx1.7=119N` mass on earth = mass on moon`=70kg`.

Discussion

2731.	SI unit of the universal constant of gravitationA. `Nm^(2)//kg^(2)`B. `Nkg^(2)//m^(2)`C. `m//s^(2)`D. `Ncm^(2)//g^(20`
Answer» Correct Answer - A `Nm^(2)//kg^(2)`

Discussion

2732.	An object weigh 10 N when measured on the surface of the earth. What would be its weight when measure on the surface of moon ?
Answer» Correct Answer - `1.67 N` `W_(e)=10N, W_(m)=?` As weight on moon `=(1)/(6)xx`weight on earth, `W_(m)=(1)/(6)xx10=1.67N`

Discussion

2733.	A man weigh 600 N on the earth. What is its mass ? Take `g=10m//s^(2)`. If he were taken on moon, his weight would be 100 N. What is his mass on moon ? What is acceleration due to gravity on moon ?
Answer» Correct Answer - `60 kg, 60 kg, 1.67 m//s^(2)` Here, `W_(e)=600N, m=?, g_(e)=10m//s^(2)` `m_(e)=W_(e)/g_(e)=600/10=60kg` mass on moon=mass on earth`=60kg` As `W_(m)=100N,` `g_(m)=W_(m)/m=100/601.67m//s^(2)`

Discussion

2734.	The SI unit of the universal constant of gravitation is ……..(a) N.m2/kg2(b) N.kg2/m2(c) m/s2(d) kg.m/s2
Answer» Answer is (a) N.m²/kg²

Discussion

2735.	How much will a person with 72 N weight on the earth, weigh on the moon? (a) 12 N (b) 36 N (c) 21 N (d) 63 N
Answer» Answer is (a) 12 N

Discussion

2736.	Name the source responsible for the motion of a planet around the Sun.
Answer» A planet revolves around the Sun due to the gravitational force exerted on it by the Sun.

Discussion

2737.	Give one example of centripetal force.
Answer» The moon revolves around the earth due to the gravitational force exerted on it by the earth. This force is directed towards the centre of the earth and is thus a centripetal force.

Discussion

2738.	What is centripetal force? (OR) Define centripetal force.
Answer» In uniform circular motion of a body, the force acting on the body is directed towards the centre of the circle. This force is called centripetal force.

Discussion

2739.	What are the factors on which the maximum height attained by a body thrown upwards depends?
Answer» The initial velocity of the body, the acceleration due to gravity at that place, the buoyant force and frictional force due to air.

Discussion

2740.	The escape velocity of a body from the earth’s surface, vsec = …….(a) \(\sqrt{\frac{GM}{R}}\)(b) \(2\sqrt{\frac{GM}{R}}\)(c) \(\sqrt{\frac{2GM}{R}}\)(d) \(\sqrt{\frac{GM}{2R}}\)
Answer» Answer is (c) \(\sqrt{\frac{2GM}{R}}\)

Discussion

2741.	What is the value of the acceleration due to gravity at the centre of the earth?
Answer» Answer is Zero.

Discussion

2742.	Write the formula for the centripetal force acting on a body performing circular motion.
Answer» Answer is F = \(\frac{mv^2}{r}\)

Discussion

2743.	Write the formula for the escape velocity of a body from the earth’s surface.
Answer» \(v_{esc}=\sqrt{\frac{2GM}{R}}\) or \(v_{esc}=\sqrt{2gR.}\)

Discussion

2744.	An artificial satellite orbiting the earth in very thin atmosphere loses its energy gradually due to small but continuous dissipation against atmospheric resistance. Then explain why its speed increases progressively as it comes closer and closer to the earth.
Answer» Correct Answer - Kinetic energy increases, but potential energy decreases, and the sum decreases due to dissipation against friction.

Discussion

2745.	Two satellite A and B, ratio of masses `3:1` are in circular orbits of radii `r and 4r`. Then ratio mechanical energy of A and B isA. `1:3`B. `3:1`C. `3:4`D. `12:1`
Answer» Correct Answer - D Energy of satellite is given by `E=-(GMm)/(2r)or E prop (m)/(r)` `rArr (E_(A))/(E_(B))=(m_(A))/(m_(B))xx(r_(B))/(r_(A))=(3)/(1)xx(4r)/(r)" "{because (m_(A))/(m_(B))=(3)/(1),(r_(B))/(r_(A))=(4r)/(r)}` `rArr (E_(A))/(E_(B))=(12)/(1)`.

Discussion

2746.	E, U and K represent total mechanical energy potential energy and kinetic energy of a satellite revolving around a planet. Which of the following quantity is not independent of orbital radius of the satellite ?A. `K+(U)/(2)`B. K + EC. `(U)/(2)+E`D. `(U)/(2)-E`
Answer» Correct Answer - C

Discussion

2747.	Two satellite A and B , ratio of masses `3 : 1` are in circular orbits of radii r and 4 r . Then ratio of total mechanical energy of A to B is
Answer» Correct Answer - `12/1`

Discussion

2748.	A satellite of mass `m` is just placed over the surface of earth. In this position mechanical energy of satellite is `E_(1)`. Now it starts orbiting round the earth in a circular path at height `h =` radius of earth. In this position, kinetic energy potential energy and total mechancial energy of satellite are `K_(2)`, `U_(2)` and `E_(2)` respectively. ThenA. `U_(2) = (E_(1))/(2)`B. `E_(2) = (E_(1))/(4)`C. `K_(2) = - E_(2)`D. `K_(2) = - (U_(2))/(2)`
Answer» Correct Answer - A::B::C::D `E_(1) = - (GMm)/(R )` `U_(2) = - (GMm)/(2R)` `(r = R + R = 2R)` `K_(2) = (GMm)/(4R)` and `E_(2) = - (GMm)/(4R)`

Discussion

2749.	A satellite of mass m is placed at a distance r from the centre of earth (mass M ). The mechanical energy of the satellite isA. `-(GMm)/(r)`B. `(GMm)/(r)`C. `(GMm)/(2r)`D. `-(GMm)/(2r)`
Answer» Correct Answer - D If a satellite of mass m is placed at a distance r form the centre of the earth (of mass M), the mechanical energy of the satellite is given by `(-GMm)/(2r)`.

Discussion

2750.	Calculate the distance from the surface of the earth at which above the surface, acceleration due to gravity is the same.
Answer» Correct Answer - `h=(sqrt(5)-1)/(2)R` Let at height or depth h from surface of earth acceleration due to gravity is same. `g_(d)=g_(h)` `g(1-(h)/(R ))=(gR^(2))/((R+h)^(2))rArr ((R-h))/(R )=(R^(2))/((R+h^(2)))` `(R-h)(R+h)^(2)=R^(3)rArr (R-h)(R^(2)+h^(2)+2hR)=R^(3)` `R^(3)+Rh^(2)+2hR^(2)-hR^(2)-h^(3)-2h^(2)R=R^(3)rArr h^(2)+Rh-R^(2)=0` `h = (-R+-sqrt(R^(2)+4R^(2)))/(2) rArr ((sqrt(5)-1)R)/(2)`

Discussion

Explore topic-wise InterviewSolutions in .

A person brings a mass of 1kg from infinity to a point A. Initially the mass was at rest but it moves at a speed of 2m/s as it reaches A. The work done by the person on the mass is —3J. The potential at A is(a) -3J/kg (b) -2J/kg (c) -5JAg (d) none of these.

A person brings a mass of 1 kg from infinity to a point . Initally the mas was at rest but it moves at a speed of 2 `ms^-1` as it reaches A. The work done by the perosn on the mass is -3J. The potential at A isA. `-3Jkg^-1`B. `-2Jkg^-1`C. `-5Jkg^-1`D. none of these

A person brings a mass of 1 kg from infinity to a point . Initally the mass was at rest but it moves at a speed of 2 `ms^-1` as it reaches A. The work done by the person on the mass is -3J. The potential at A isA. `-3 "J kg"^(-1)`B. `-12 "J kg"^(-1)`C. `-5 "J kg"^(-1)`D. None of these

A satallite of mass `m`, initally at rest on the earth, is launched into a circular orbit at a height equal to the the radius of the earth. The minimum energy required isA. `(sqrt(3))/4 mgR`B. `1/2 mgR`C. `1/4 mgR`D. `3/4 mgR`

A satallite of mass `m`, initally at rest on the earth, is launched into a circular orbit at a height equal to the the radius of the earth. The minimum energy required isA. `(sqrt(3))/(4)mgR`B. `(1)/(2)mgR`C. `(1)/(4)mgR`D. `(3)/(4)mgR`

A communications Earth satalliteA. goes round the earth from east to westB. can be in the equatorial plane onlyC. can be vertically any place on the earthD. goes round the earth from west to east

A bullet is fired vertically upwards with velocity v form the surface of a spherical planet. If the escape velocity from the planet is `v_(esc) = vsqrtN`, then the value of N is (ignore energy loss due to atmosphere)

A communications Earth satalliteA. goes round the earth from east to westB. can be in the equatorial plane onlyC. can be vertically above any place on the earthD. goes round the earth from west to east

Two particles of masses `1kg` and `2kg` are placed at a distance of `50cm`. Find the initial acceleration of the first particle due to gravitational force.

For particles of equal masses M that move along a circle of radius R under the action of their mutual gravitational attraction. Find the speed of each particle.

Two particles of equal mass (m) each move in a circle of radius (r) under the action of their mutual gravitational attraction find the speed of each particle.

The gravitational potential due to the Earth is minimum at (A) the centre of the Earth. (B) the surface of the Earth. (C) a points inside the Earth but not at its centre. (D) infinite distance.

State Kepler’s law equal of area.

The binding energy of a satellite revolving around planet in a circular orbit is 3 × 109 J. Its kinetic energy is ______ (A) 6 × 109 J (B) -3 × 109 J (C) -6 × 10+9 J (D) 3 × 10+9 J

In case of an orbiting satellite, if the radius of orbit is decreasedA. its Kinetic Energy decreasesB. its Potential Energy decreasesC. its Mechanical Energy decreasesD. its speed decreases

How much energy is required by a satellite to keep it orbiting? Neglect air resistance, why?

Gravitational force between a point mass m and M separated by a distance is F. Now if a point mass 2m is placed next to m is contact with it. The force on M due to `m` and the total force on M areA. 2F, FB. F, 2FC. F, 3FD. F, F

The distance of planet Jupiter from the Sun is `5.2` times that of the earth. Find the period of revolution of Jupiter around the Sun.

Gravitational force between two point masses `m` and `M` separated by a distance `r` is `F`. Now if a point mass `3m` is placed next to `m`, what will be that a. force on `M` due to `m`, b total on `M`?

A ball is thrown vertically upwards with a velocity of `98m//s`. Calculate (i) The maximum height to which it rises. (ii) Total time it takes to return to earth.

How much will a body of mass once kg weigh on moon ? Given mass of moon is `(1)/(100)` mass of earth and diameter of moon is `(1)/(4)` the diameter of earth.

How much would a 70 kg man weigh on moon ?what will be his mass on earth and moon ? Given g on moon`=1.7 m//s^(2)`.

SI unit of the universal constant of gravitationA. `Nm^(2)//kg^(2)`B. `Nkg^(2)//m^(2)`C. `m//s^(2)`D. `Ncm^(2)//g^(20`

An object weigh 10 N when measured on the surface of the earth. What would be its weight when measure on the surface of moon ?

A man weigh 600 N on the earth. What is its mass ? Take `g=10m//s^(2)`. If he were taken on moon, his weight would be 100 N. What is his mass on moon ? What is acceleration due to gravity on moon ?

The SI unit of the universal constant of gravitation is ……..(a) N.m2/kg2(b) N.kg2/m2(c) m/s2(d) kg.m/s2

How much will a person with 72 N weight on the earth, weigh on the moon? (a) 12 N (b) 36 N (c) 21 N (d) 63 N

Name the source responsible for the motion of a planet around the Sun.

Give one example of centripetal force.

What is centripetal force? (OR) Define centripetal force.

What are the factors on which the maximum height attained by a body thrown upwards depends?

The escape velocity of a body from the earth’s surface, vsec = …….(a) \(\sqrt{\frac{GM}{R}}\)(b) \(2\sqrt{\frac{GM}{R}}\)(c) \(\sqrt{\frac{2GM}{R}}\)(d) \(\sqrt{\frac{GM}{2R}}\)

What is the value of the acceleration due to gravity at the centre of the earth?

Write the formula for the centripetal force acting on a body performing circular motion.

Write the formula for the escape velocity of a body from the earth’s surface.

An artificial satellite orbiting the earth in very thin atmosphere loses its energy gradually due to small but continuous dissipation against atmospheric resistance. Then explain why its speed increases progressively as it comes closer and closer to the earth.

Two satellite A and B, ratio of masses `3:1` are in circular orbits of radii `r and 4r`. Then ratio mechanical energy of A and B isA. `1:3`B. `3:1`C. `3:4`D. `12:1`

E, U and K represent total mechanical energy potential energy and kinetic energy of a satellite revolving around a planet. Which of the following quantity is not independent of orbital radius of the satellite ?A. `K+(U)/(2)`B. K + EC. `(U)/(2)+E`D. `(U)/(2)-E`

Two satellite A and B , ratio of masses `3 : 1` are in circular orbits of radii r and 4 r . Then ratio of total mechanical energy of A to B is

A satellite of mass m is placed at a distance r from the centre of earth (mass M ). The mechanical energy of the satellite isA. `-(GMm)/(r)`B. `(GMm)/(r)`C. `(GMm)/(2r)`D. `-(GMm)/(2r)`

Calculate the distance from the surface of the earth at which above the surface, acceleration due to gravity is the same.